Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $n = \dfrac{6t - 6}{-3t + 30} \div \dfrac{t^2 + 5t - 6}{-8t - 48} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{6t - 6}{-3t + 30} \times \dfrac{-8t - 48}{t^2 + 5t - 6} $ First factor the quadratic. $n = \dfrac{6t - 6}{-3t + 30} \times \dfrac{-8t - 48}{(t - 1)(t + 6)} $ Then factor out any other terms. $n = \dfrac{6(t - 1)}{-3(t - 10)} \times \dfrac{-8(t + 6)}{(t - 1)(t + 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 6(t - 1) \times -8(t + 6) } { -3(t - 10) \times (t - 1)(t + 6) } $ $n = \dfrac{ -48(t - 1)(t + 6)}{ -3(t - 10)(t - 1)(t + 6)} $ Notice that $(t + 6)$ and $(t - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -48\cancel{(t - 1)}(t + 6)}{ -3(t - 10)\cancel{(t - 1)}(t + 6)} $ We are dividing by $t - 1$ , so $t - 1 \neq 0$ Therefore, $t \neq 1$ $n = \dfrac{ -48\cancel{(t - 1)}\cancel{(t + 6)}}{ -3(t - 10)\cancel{(t - 1)}\cancel{(t + 6)}} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $n = \dfrac{-48}{-3(t - 10)} $ $n = \dfrac{16}{t - 10} ; \space t \neq 1 ; \space t \neq -6 $